4 and 2 electrons are present.
Answer: The concentration of 
ions in vinegar is 0.001 M.
Explanation:
Given: pH = 3.0
pH is the negative logarithm of concentration of hydrogen ion.
The expression for pH is as follows.
![pH = - log [H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D)
Substitute the value into above expression as follows.
![pH = - log [H^{+}]\\3.0 = - log [H^{+}]\\conc. of H^{+} = antilog (- 3.0)\\= 0.001 M](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5C3.0%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5Cconc.%20of%20H%5E%7B%2B%7D%20%3D%20antilog%20%28-%203.0%29%5C%5C%3D%200.001%20M)
Thus, we can conclude that the concentration of ions in vinegar is 0.001 M.
Answer:
CH4
Explanation:
The number of moles of carbon and hydrogen has been given as follows:
C = 0.300 mol
H = 1.20 mol
Next, we divide each mole value by the smallest (0.300)
C = 0.300 ÷ 0.300 = 1
H = 1.20 ÷ 0.300 = 4
The empirical ratio of Carbon and Hydrogen is 1:4, hence, the empirical formula is CH4
The net ionic equation of the reaction could be determined by cancelling out the like ions between both sides of the reaction. These ions are called spectator ions. They are called as such because they do not actively participate in the reaction. The spectator ions are Na+ and Cl-. When you cancel those, the equation would become letter D.
